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Here is a table summarizing the logarithmic properties/rules/identities for any \(b>0\), and \(x,y\in\mathbb{R}\). We will go through each of them in depth in the below sections.

Switching between Logarithmic and Exponential Form |
\(y=b^x \iff log_b(y)=x\) |

Product Rule |
\(log_b(x\cdot y)=log_b(x)+log_b(y)\) |

Quotient Rule |
\(log_b(\frac{x}{y})=log_b(x)-log_b(y)\) |

Power Rule |
\(log_b(x^n)=n \cdot log_b(x)\) |

Change of Base |
\(log_b(x)=\frac{log_a(x)}{log_a(b)}\), for any \(a\in\mathbb{R}\) |

Cancelling Exponential Identity #1 |
\(b^{log_b(x)}=x\) |

Cancelling Exponential Identity #2 |
\(log_b(b^x)=x\) |

Logarithm of the Base |
\(log_b(b)=1\) |

Logarithm of 1 |
\(log_b(1)=0\) |

__Notes__

- You may find other properties introduced in your course notes or textbook, but they will most likely be derived from one (or more) of the above properties/rules/identities.
- We covered how to switch between logarithmic and exponential form on the Converting between Exponential and Logarithmic Form page, so we won't cover it below. Please visit that page to see examples and find practice questions for this topic.

Recall that the product and quotient rules for logarithms are:

\(log_b(x\cdot y)=log_b(x)+log_b(y)\)

\(log_b(\frac{x}{y})=log_b(x)-log_b(y)\)

Since we can think of logarithms as exponents, we can actually derive these rules with the help of the product/quotient rules for exponents:

\(b^m \cdot b^n=b^{m+n}\)

\(\frac{b^m}{b^n}=b^{m-n}\)

For the product rule, suppose we let \(x=b^m\) and \(y=b^n\) for some \(m,n\in\mathbb{R}\). Then, switching from exponential to logarithmic form gives us:

\(m=log_b(x)\) and \(n=log_b(y)\)

So we have:

\(log_b(x\cdot y)\)

\(=log_b((b^m)(b^n))\) by substituting in \(x=b^m\) and \(y=b^n\)

\(=log_b(b^{m+n})\) by applying the exponent product rule

\(=m+n\) by using the "Cancelling Exponential Identity #2"

\(=log_b(x)+log_b(y)\) by substituting in \(m=log_b(x)\) and \(n=log_b(y)\)

*__Note__: The quotient rule for logarithms can be derived using similar steps, but with the help of the exponent quotient rule.

Example: Write the expression \(log_4(15)\) as the sum of two logarithms.

**Solution:**

Since \(15=3\cdot 5\), we have \(log_4(15)=log_4(3\cdot 5)\). Then, applying the product rule gives:

\(log_4(15)=log_4(3\cdot 5)=log_4(3)+log_4(5)\)

Example: Write the expression \(log_{11}(14)-log_{11}(2)\) as a single logarithm, and simplify if possible.

**Solution:**

Applying the quotient rule to the expression gives:

\(log_{11}(14)-log_{11}(2)=log_{11}(\frac{14}{2})=log_{11}(7)\)

The power rule for logarithms is:

\(log_b(x^n)=n\cdot log_b(x)\)

Similar to the product and quotient rules, we can use the power rule for exponents to derive the power rule for logarithms.

Recall that the power rule for exponents is:

\((b^n)^m=b^{n\cdot m}\)

To derive the power rule for logarithms, suppose we let \(x=b^m\), so we have \(m=log_b(x)\). So starting with \(log_b(x^n)\), we have:

\(log_b(x^n)\)

\(=log_b((b^m)^n\) by substituting in \(x=b^m\)

\(=log_b(b^{m\cdot n}\) by applying power rule for exponents

\(=mn\) by using "Cancelling Exponential Identity #2"

\(=n\cdot log_b(x)\) by substituting in \(m=log_b(x)\)

Example: Evaluate the expression \(log_3(27^2)\) using the power rule for logarithms.

**Solution:**

Applying the power rule to this expression gives:

\(log_3(27^2)=2\cdot log_3(27)\)

Since \(log_3(27)=3\), we have the following calculation:

\(log_3(27^2)=2\cdot log_3(27)= 2\cdot 3 = 6\)

For most of our calculators, we are only able to calculate logs of base 10 or base \(e\)* *(see Common and Natural Logarithm to review). Fortunately, it is possible to change the base of logarithms with the help of the Change of Base formula:

\(log_b(x)=\frac{log_a(x)}{log_a(b)}\), for any \(a \in \mathbb{R}\)

The \(a\) value is the new base of our choice, which will typically be either 10 or \(e\) (so we can evaluate log expressions using our calculators).

Example: Determine the approximate value of \(log_4(17)\) using your calculator by changing the base to 10 and \(e\), rounded to 3 decimal places.

**Solution:**

Applying the change of base formula gives:

Base \(10\) (Common Logarithm) |
Base \(e\) (Natural Logarithm) |

\(log_4(17)\) \(=\frac{log_{10}(17)}{log_{10}(4)}\) \(=\frac{log(17)}{log(4)}\) \(= 2.044\) |
\(log_4(17)\) \(=\frac{log_e(17)}{log_e(4)}\) \(=\frac{ln(17)}{ln(4)}\) \(=2.044\) |

*__Note__: As seen in this example, regardless of the new base you pick, the final value will be the same.

The 4 identities included in the above summary table were:

Cancelling Exponential Identity #1 |
\(b^{log_b(x)}=x\) |

Cancelling Exponential Identity #2 |
\(log_b(b^x)=x\) |

Logarithm of the Base |
\(log_b(b)=1\) |

Logarithm of 1 |
\(log_b(1)=0\) |

All these property/rules can all be derived straight from the definitions of logarithm and exponent, and the fact that they're inverses:

\(x=b^y \iff y=log_b(x)\)

1. __Cancelling Exponential Identity #1__: Given \(x=b^y\), we can substitute in \(y=log_b(x)\) to get \(x=b^{log_b(x)}\).

2. __Cancelling Exponential Identity #2__: Given \(y=log_b(x)\), we can substitute in \(x=b^y\) to get \(y=log_b(b^y)\).

3. __Logarithm of the Base__: \(log_b(b)=1\) since \(b^1=b\), for any \(b \in \mathbb{R}\).

4. __Logarithm of 1__: \(log_b(1)=0\) since \(b^0=1\), for any \(b \in \mathbb{R}\).

Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.

- Last Updated: Aug 2, 2022 11:43 AM
- URL: https://libraryguides.centennialcollege.ca/c.php?g=718098
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